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sudhanshu746

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artofproblemsolving

like 0

e65442156206126c5b8ecd5a38245cb4

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_2

How many different combinations of $5 bills and $2 bills can be used to make a total of $17? Order does not matter in this problem. $\text {(A)}\ 2 \qquad \text {(B)}\ 3 \qquad \text {(C)}\ 4 \qquad \text {(D)}\ 5 \qquad \text {(E)}\ 6$

You cannot use more than 4 $<dollar>5$ bills, but if you use 3 $<dollar>5$ bills, you can add another $<dollar>2$ bill to make a combination. You cannot use 2 $<dollar>5$ bills since you have an odd number of dollars that need to be paid with $<dollar>2$ bills. You can also use 1 $<dollar>5$ bill and 6 $<dollar>2$ bills to make another combination. There are no other possibilities, as making $<dollar>17$ with 0 $<dollar>5$ bills is impossible, so the answer is $\boxed{2}$

A

2

b933f9b6f2d4170c37f408b5370a2dc7

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_3

What is the smallest possible average of four distinct positive even integers? $\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7$

In order to get the smallest possible average, we want the 4 even numbers to be as small as possible. The first 4 positive even numbers are 2, 4, 6, and 8. Their average is $\frac{2+4+6+8}{4}=\boxed{5}$

C

5

69cc946117a2a70fd0f3ae3e5f81143d

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_4

The year 2002 is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after 2002 that is a palindrome? $\text{(A)}\ 0 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 25$

The palindrome right after 2002 is 2112. The product of the digits of 2112 is $\boxed{4}$

B

4

69cc946117a2a70fd0f3ae3e5f81143d

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_4

The year 2002 is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after 2002 that is a palindrome? $\text{(A)}\ 0 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 25$

The palindrome formula is to add 110 to the number in order to get the next palindrome, a palindrome needs to be in the form as ABBA . We can use this in this case to get 2112. 2*1*1*2=4. Therefore, the answer is $\boxed{4}$

B

4

2b3374033e644b27bfd9c80b0de7e04d

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_7

The students in Mrs. Sawyer's class were asked to do a taste test of five kinds of candy. Each student chose one kind of candy. A bar graph of their preferences is shown. What percent of her class chose candy E? [asy] real[] r={6, 8, 4, 2, 5}; int i; for(i=0; i<5; i=i+1) { filldraw((4i,0)--(4i+3,0)--(4i+3,2r[i])--(4i,2r[i])--cycle, black, black); } draw(origin--(19,0)--(19,16)--(0,16)--cycle, linewidth(0.9)); for(i=1; i<8; i=i+1) { draw((0,2i)--(19,2i)); } label("$0$", (0,2*0), W); label("$1$", (0,2*1), W); label("$2$", (0,2*2), W); label("$3$", (0,2*3), W); label("$4$", (0,2*4), W); label("$5$", (0,2*5), W); label("$6$", (0,2*6), W); label("$7$", (0,2*7), W); label("$8$", (0,2*8), W); label("$A$", (0*4+1.5, 0), S); label("$B$", (1*4+1.5, 0), S); label("$C$", (2*4+1.5, 0), S); label("$D$", (3*4+1.5, 0), S); label("$E$", (4*4+1.5, 0), S); label("SWEET TOOTH", (9.5,18), N); label("Kinds of candy", (9.5,-2), S); label(rotate(90)*"Number of students", (-2,8), W);[/asy] $\text{(A)}\ 5 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20$

From the bar graph, we can see that $5$ students chose candy E. There are $6+8+4+2+5=25$ total students in Mrs. Sawyers class. The percent that chose E is $\frac{5}{25} \cdot 100 = \boxed{20}$

E

20

16dffb75f3e15c4823f6db7d163398ed

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_8

Problems 8,9 and 10 use the data found in the accompanying paragraph and table: [asy] /* AMC8 2002 #8, 9, 10 Problem */ size(3inch, 1.5inch); for ( int y = 0; y &lt;= 5; ++y ) { draw((0,y)--(18,y)); } draw((0,0)--(0,5)); draw((6,0)--(6,5)); draw((9,0)--(9,5)); draw((12,0)--(12,5)); draw((15,0)--(15,5)); draw((18,0)--(18,5)); draw(scale(0.8)*"50s", (7.5,4.5)); draw(scale(0.8)*"4", (7.5,3.5)); draw(scale(0.8)*"8", (7.5,2.5)); draw(scale(0.8)*"6", (7.5,1.5)); draw(scale(0.8)*"3", (7.5,0.5)); draw(scale(0.8)*"60s", (10.5,4.5)); draw(scale(0.8)*"7", (10.5,3.5)); draw(scale(0.8)*"4", (10.5,2.5)); draw(scale(0.8)*"4", (10.5,1.5)); draw(scale(0.8)*"9", (10.5,0.5)); draw(scale(0.8)*"70s", (13.5,4.5)); draw(scale(0.8)*"12", (13.5,3.5)); draw(scale(0.8)*"12", (13.5,2.5)); draw(scale(0.8)*"6", (13.5,1.5)); draw(scale(0.8)*"13", (13.5,0.5)); draw(scale(0.8)*"80s", (16.5,4.5)); draw(scale(0.8)*"8", (16.5,3.5)); draw(scale(0.8)*"15", (16.5,2.5)); draw(scale(0.8)*"10", (16.5,1.5)); draw(scale(0.8)*"9", (16.5,0.5)); label(scale(0.8)*"Country", (3,4.5)); label(scale(0.8)*"Brazil", (3,3.5)); label(scale(0.8)*"France", (3,2.5)); label(scale(0.8)*"Peru", (3,1.5)); label(scale(0.8)*"Spain", (3,0.5)); label(scale(0.9)*"Juan's Stamp Collection", (9,0), S); label(scale(0.9)*"Number of Stamps by Decade", (9,5), N);[/asy] How many of his European stamps were issued in the '80s? $\text{(A)}\ 9 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 42$

France and Spain are European countries. The number of '80s stamps from France is $15$ and the number of '80s stamps from Spain is $9$ . The total number of stamps is $15+9=\boxed{24}$

D

24

ac1a24b9bcb728d734f04b1193e5266d

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_9

Problems 8,9 and 10 use the data found in the accompanying paragraph and table: [asy] /* AMC8 2002 #8, 9, 10 Problem */ size(3inch, 1.5inch); for ( int y = 0; y &lt;= 5; ++y ) { draw((0,y)--(18,y)); } draw((0,0)--(0,5)); draw((6,0)--(6,5)); draw((9,0)--(9,5)); draw((12,0)--(12,5)); draw((15,0)--(15,5)); draw((18,0)--(18,5)); draw(scale(0.8)*"50s", (7.5,4.5)); draw(scale(0.8)*"4", (7.5,3.5)); draw(scale(0.8)*"8", (7.5,2.5)); draw(scale(0.8)*"6", (7.5,1.5)); draw(scale(0.8)*"3", (7.5,0.5)); draw(scale(0.8)*"60s", (10.5,4.5)); draw(scale(0.8)*"7", (10.5,3.5)); draw(scale(0.8)*"4", (10.5,2.5)); draw(scale(0.8)*"4", (10.5,1.5)); draw(scale(0.8)*"9", (10.5,0.5)); draw(scale(0.8)*"70s", (13.5,4.5)); draw(scale(0.8)*"12", (13.5,3.5)); draw(scale(0.8)*"12", (13.5,2.5)); draw(scale(0.8)*"6", (13.5,1.5)); draw(scale(0.8)*"13", (13.5,0.5)); draw(scale(0.8)*"80s", (16.5,4.5)); draw(scale(0.8)*"8", (16.5,3.5)); draw(scale(0.8)*"15", (16.5,2.5)); draw(scale(0.8)*"10", (16.5,1.5)); draw(scale(0.8)*"9", (16.5,0.5)); label(scale(0.8)*"Country", (3,4.5)); label(scale(0.8)*"Brazil", (3,3.5)); label(scale(0.8)*"France", (3,2.5)); label(scale(0.8)*"Peru", (3,1.5)); label(scale(0.8)*"Spain", (3,0.5)); label(scale(0.9)*"Juan's Stamp Collection", (9,0), S); label(scale(0.9)*"Number of Stamps by Decade", (9,5), N);[/asy] In dollars and cents, how much did his South American stamps issued before the ’70s cost him? $\text{(A)}\ $0.40 \qquad \text{(B)}\ $1.06 \qquad \text{(C)}\ $1.80 \qquad \text{(D)}\ $2.38 \qquad \text{(E)}\ $2.64$

Brazil 50s and 60s total 11 stamps with each 6 cents, Peru 50s and 60s total 10 stamps with each 4 cents. So total $11*0.06+10*0.04 = \boxed{1.06}$

B

1.06

f7aac714a9a971d023b425e78fb13b4e

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_10

Problems 8,9 and 10 use the data found in the accompanying paragraph and table: The average price of his '70s stamps is closest to $\text{(A)}\ 3.5 \text{ cents} \qquad \text{(B)}\ 4 \text{ cents} \qquad \text{(C)}\ 4.5 \text{ cents} \qquad \text{(D)}\ 5 \text{ cents} \qquad \text{(E)}\ 5.5 \text{ cents}$

The price of all the stamps in the '70s together over the total number of stamps is equal to the average price. \[\frac{(12)(0.06)+(12)(0.06)+(6)(0.04)+(13)(0.05)}{12+12+6+13}\\ = \frac{0.72+0.72+0.24+0.65}{43}\\ = \frac{2.33}{43} \approx \boxed{5.5}\]

E

5.5

cad911729b80c8add923db40434bca12

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_11

A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth? [asy] path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(3,0)*p); draw(shift(3,1)*p); draw(shift(4,0)*p); draw(shift(4,1)*p); draw(shift(7,0)*p); draw(shift(7,1)*p); draw(shift(7,2)*p); draw(shift(8,0)*p); draw(shift(8,1)*p); draw(shift(8,2)*p); draw(shift(9,0)*p); draw(shift(9,1)*p); draw(shift(9,2)*p);[/asy] $\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

The first square has a sidelength of $1$ , the second square $2$ , and so on. The seventh square has $7$ and is made of $7^2=49$ unit tiles. The sixth square has $6$ and is made of $6^2=36$ unit tiles. The seventh square has $49-36=\boxed{13}$ more tiles than the sixth square.

C

13

cad911729b80c8add923db40434bca12

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_11

A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth? [asy] path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(3,0)*p); draw(shift(3,1)*p); draw(shift(4,0)*p); draw(shift(4,1)*p); draw(shift(7,0)*p); draw(shift(7,1)*p); draw(shift(7,2)*p); draw(shift(8,0)*p); draw(shift(8,1)*p); draw(shift(8,2)*p); draw(shift(9,0)*p); draw(shift(9,1)*p); draw(shift(9,2)*p);[/asy] $\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

The edge of each square is one tile length longer than the edge of the previous square, which means that each square has $2*$ edge length $- 1$ more tiles than the previous square, because each square is just one edge added on the top and on the right to the previous square, with one overlapping tile. Then the seventh square has $2*7-1=13$ more tiles than the sixth square, which is $\boxed{13}$

C

13

cad911729b80c8add923db40434bca12

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_11

A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth? [asy] path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(3,0)*p); draw(shift(3,1)*p); draw(shift(4,0)*p); draw(shift(4,1)*p); draw(shift(7,0)*p); draw(shift(7,1)*p); draw(shift(7,2)*p); draw(shift(8,0)*p); draw(shift(8,1)*p); draw(shift(8,2)*p); draw(shift(9,0)*p); draw(shift(9,1)*p); draw(shift(9,2)*p);[/asy] $\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

We see a pattern of 1, 4, and 9, all of which are the squares of 1, 2,and 3 respectively. So, the 6th and 7th squares will also follow the same pattern. Via the difference of squares, we see that $7^2 - 6^2$ . Now we can see that the seventh square has $(7-6)(7+6) =$ $\boxed{13}$ more tiles than the sixth square.

C

13

56a18605a539f3d9b5d89f89a66aba05

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_12

A board game spinner is divided into three regions labeled $A$ $B$ and $C$ . The probability of the arrow stopping on region $A$ is $\frac{1}{3}$ and on region $B$ is $\frac{1}{2}$ . The probability of the arrow stopping on region $C$ is: $\text{(A)}\ \frac{1}{12}\qquad\text{(B)}\ \frac{1}{6}\qquad\text{(C)}\ \frac{1}{5}\qquad\text{(D)}\ \frac{1}{3}\qquad\text{(E)}\ \frac{2}{5}$

Since the arrow must land in one of the three regions, the sum of the probabilities must be 1. Thus the answer is $1-\frac{1}{2}-\frac{1}{3}=\boxed{16}$

B

16

6482e313c16dc63509616e4bba60d81c

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_13

For his birthday, Bert gets a box that holds 125 jellybeans when filled to capacity. A few weeks later, Carrie gets a larger box full of jellybeans. Her box is twice as high, twice as wide and twice as long as Bert's. Approximately, how many jellybeans did Carrie get? $\text{(A)}\ 250\qquad\text{(B)}\ 500\qquad\text{(C)}\ 625\qquad\text{(D)}\ 750\qquad\text{(E)}\ 1000$

Since the volume ratio is equal to the sides ratio cubed, then the ratio of the larger box's volume to the smaller one is 2 cubed. $2^3=8$ Now multiply 125 (the number of jellybeans that Bert's box can hold) by 8. $8\cdot125= \boxed{1000}$

E

1000

a82a37455bae633cd86f574c5d1b7927

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_14

A merchant offers a large group of items at $30\%$ off. Later, the merchant takes $20\%$ off these sale prices. The total discount is $\text{(A)}\ 35\%\qquad\text{(B)}\ 44\%\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\%$

Let's assume that each item is $100$ dollars. First we take off $30\%$ off of $100$ dollars. $100\cdot0.7=70$ Next, we take off the extra $20\%$ as asked by the problem. $70\cdot0.80=56$ So the final price of an item is $56. We have to do $100-56$ because $56$ was the final price and we wanted the discount. So the final answer is $44\%$ , which is answer choice $\boxed{44}$

B

44

6550a90f21f2323e2280b862e97afd45

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_17

In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have? $\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$

We can try to guess and check to find the answer. If she got five right, her score would be $(5*5)-(5*2)=15$ . If she got six right her score would be $(6*5)-(2*4)=22$ . That's close, but it's still not right! If she got 7 right, her score would be $(7*5)-(2*3)=29$ . Thus, our answer is $\boxed{7}$ . ~avamarora

C

7

6550a90f21f2323e2280b862e97afd45

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_17

In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have? $\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$

We can start with the full score, 50, and subtract not only 2 points for each incorrect answer but also the 5 points we gave her credit for. This expression is equivalent to her score, 29. Let $x$ be the number of questions she answers correctly. Then, we will represent the number incorrect by $10-x$ \begin{align*} 50-7(10-x)&=29\\ 50-70+7x&=29\\ 7x&=49\\ x&=\boxed{7}

C

7

6550a90f21f2323e2280b862e97afd45

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_17

In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have? $\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$

Suppose she got $x$ questions right. Then she got $10 - x$ questions wrong. Since she gains 5 points for a correct answer and loses 2 for an incorrect one, we can solve $5x - 2(10 - x) = 29$ to get that $x = \boxed{7}$

C

7

f4c22c651118bdc44e0967f62a725bfc

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_18

Gage skated $1$ hr $15$ min each day for $5$ days and $1$ hr $30$ min each day for $3$ days. How long would he have to skate the ninth day in order to average $85$ minutes of skating each day for the entire time? $\text{(A)}\ \text{1 hr}\qquad\text{(B)}\ \text{1 hr 10 min}\qquad\text{(C)}\ \text{1 hr 20 min}\qquad\text{(D)}\ \text{1 hr 40 min}\qquad\text{(E)}\ \text{2 hr}$

Converting into minutes and adding, we get that she skated $75*5+90*3+x = 375+270+x = 645+x$ minutes total, where $x$ is the amount she skated on day $9$ . Dividing by $9$ to get the average, we get $\frac{645+x}{9}=85$ . Solving for $x$ \[645+x=765\] \[x=120\] Now we convert back into hours and minutes to get $\boxed{2}$

E

2

f4c22c651118bdc44e0967f62a725bfc

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_18

Gage skated $1$ hr $15$ min each day for $5$ days and $1$ hr $30$ min each day for $3$ days. How long would he have to skate the ninth day in order to average $85$ minutes of skating each day for the entire time? $\text{(A)}\ \text{1 hr}\qquad\text{(B)}\ \text{1 hr 10 min}\qquad\text{(C)}\ \text{1 hr 20 min}\qquad\text{(D)}\ \text{1 hr 40 min}\qquad\text{(E)}\ \text{2 hr}$

For the first five days, each day you are $10$ minutes short of $85$ minutes. And for the next three days, you are $5$ minutes above $85$ minutes. So in total you are missing $3*5-5*10$ , which equals to negative $35$ . So on the ninth day, to have an average of $85$ minutes, Gage need to skate for $85+35$ minutes, which is $120$ minutes, or $\boxed{2}$

E

2

acadecfe184db1bd5c1c600851ffa10e

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_19

How many whole numbers between 99 and 999 contain exactly one 0? $\text{(A)}\ 72\qquad\text{(B)}\ 90\qquad\text{(C)}\ 144\qquad\text{(D)}\ 162\qquad\text{(E)}\ 180$

Numbers with exactly one zero have the form $\overline{a0b}$ or $\overline{ab0}$ , where the $a,b \neq 0$ . There are $(9\cdot1\cdot9)+(9\cdot9\cdot1) = 81+81 = \boxed{162}$

D

162

db320b40ed9600084b98092bae8ee960

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_20

The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$ . Altitude $\overline{XC}$ bisects $\overline{YZ}$ . The area (in square inches) of the shaded region is [asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); draw((5,4)--(5,0)); fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); label(scale(0.8)*"$X$", (5,4), N); label(scale(0.8)*"$Y$", (0,0), W); label(scale(0.8)*"$Z$", (10,0), E); label(scale(0.8)*"$A$", (2.5,2.2), W); label(scale(0.8)*"$B$", (7.5,2.2), E); label(scale(0.8)*"$C$", (5,0), S); fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white);[/asy] $\textbf{(A)}\ 1\frac{1}2\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2\frac{1}2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 3\frac{1}2$

The shaded region is a right trapezoid. Assume WLOG that $YZ=8$ . Then because the area of $\triangle XYZ$ is equal to 8, the height of the triangle $XC=2$ . Because the line $AB$ is a midsegment, the top base of the trapezoid is $\frac12 AB = \frac14 YZ = 2$ . Also, $AB$ divides $XC$ in two, so the height of the trapezoid is $\frac12 (2) = 1$ . The bottom base is $\frac12 YZ = 4$ . The area of the shaded region is $\frac12 (2+4)(1) = \boxed{3}$

D

3

db320b40ed9600084b98092bae8ee960

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_20

The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$ . Altitude $\overline{XC}$ bisects $\overline{YZ}$ . The area (in square inches) of the shaded region is [asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); draw((5,4)--(5,0)); fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); label(scale(0.8)*"$X$", (5,4), N); label(scale(0.8)*"$Y$", (0,0), W); label(scale(0.8)*"$Z$", (10,0), E); label(scale(0.8)*"$A$", (2.5,2.2), W); label(scale(0.8)*"$B$", (7.5,2.2), E); label(scale(0.8)*"$C$", (5,0), S); fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white);[/asy] $\textbf{(A)}\ 1\frac{1}2\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2\frac{1}2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 3\frac{1}2$

Since $A$ and $B$ are the midpoints of $XY$ and $XZ$ , respectively, $AY=AX=BX=BZ$ . Draw segments $AC$ and $BC$ . Drawing an altitude in an isoceles triangle splits the triangle into 2 congruent triangles and we also know that $YC=CZ$ $AB$ is the line that connects the midpoints of two sides of a triangle together, which means that $AB$ is parallel to and half in length of $YZ$ . Then $AB=YC=CZ$ . Since $AB$ is parallel to $YZ$ , and $XY$ is the transversal, $\angle XAB=\angle AYC.$ Similarly, $\angle XBA=\angle BZC.$ Then, by SAS, $\triangle XAB=\triangle AYC=\triangle BZC$ . Since corresponding parts of congruent triangles are congruent, $AC=BC=XA$ . Since ACY and BCZ are now isosceles triangles, $\angle AYC=\angle ACY$ and $\angle BZC=\angle BCZ$ Using the fact that $AB$ is parallel to $YZ$ $\angle ACY=\angle CAB$ and $\angle BCZ=\angle CBA$ . Now $\triangle XAB=\triangle AYC=\triangle BZC=\triangle ABC$ . Draw an altitude through each of them such that each triangle is split into two congruent right triangles. Now there are a total of 8 congruent small triangles, each with area 1. The shaded area has three of these triangles, so it has area 3. Basically the proof is to show $\triangle XAB=\triangle AYC=\triangle BZC=\triangle ABC$ . If you just look at the diagram you can easily see that the triangles are congruent and you would solve this a lot faster. Anyways, since those triangles are congruent, you can split each in half to find eight congruent triangles with area 1, and since the shaded region has three of these triangles, its area is $\boxed{3}$

D

3

db320b40ed9600084b98092bae8ee960

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_20

The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$ . Altitude $\overline{XC}$ bisects $\overline{YZ}$ . The area (in square inches) of the shaded region is [asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); draw((5,4)--(5,0)); fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); label(scale(0.8)*"$X$", (5,4), N); label(scale(0.8)*"$Y$", (0,0), W); label(scale(0.8)*"$Z$", (10,0), E); label(scale(0.8)*"$A$", (2.5,2.2), W); label(scale(0.8)*"$B$", (7.5,2.2), E); label(scale(0.8)*"$C$", (5,0), S); fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white);[/asy] $\textbf{(A)}\ 1\frac{1}2\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2\frac{1}2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 3\frac{1}2$

We know the area of triangle $XYZ$ is $8$ square inches. The area of a triangle can also be represented as $\frac{bh}{2}$ or in this problem $\frac{XC\cdot YZ}{2}$ . By solving, we have \[\frac{XC\cdot YZ}{2} = 8,\] \[XC\cdot YZ = 16.\] With SAS congruence, triangles $XCY$ and $XCZ$ are congruent. Hence, triangle $XCY = \frac{8}{2} = 4$ . (Let's say point $D$ is the intersection between line segments $XC$ and $AB$ .) We can find the area of the trapezoid $ADCY$ by subtracting the area of triangle $XAD$ from $4$ We find the area of triangle $XAD$ by the $\frac{bh}{2}$ formula- $\frac{XD\cdot AD}{2} = \frac{\frac{XC}{2}\cdot AD}{2}$ $AD$ is $\frac{1}{4}$ of $YZ$ from solution 1. The area of $XAD$ is \[\frac{\frac{XC}{2}\cdot \frac{YZ}{4}}{2} = \frac{16}{16} = 1\] Therefore, the area of the shaded area- trapezoid $ADCY$ has area $4-1 = \boxed{3}$

D

3

db320b40ed9600084b98092bae8ee960

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_20

The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$ . Altitude $\overline{XC}$ bisects $\overline{YZ}$ . The area (in square inches) of the shaded region is [asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); draw((5,4)--(5,0)); fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); label(scale(0.8)*"$X$", (5,4), N); label(scale(0.8)*"$Y$", (0,0), W); label(scale(0.8)*"$Z$", (10,0), E); label(scale(0.8)*"$A$", (2.5,2.2), W); label(scale(0.8)*"$B$", (7.5,2.2), E); label(scale(0.8)*"$C$", (5,0), S); fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white);[/asy] $\textbf{(A)}\ 1\frac{1}2\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2\frac{1}2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 3\frac{1}2$

$\usepackage[dvipsnames]{xcolor} \textcolor{BlueViolet}{\text{Super fast after convincing yourself they are congruent}}$ [asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); draw((5,4)--(5,0)); fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); label(scale(0.8)*"$X$", (5,4), N); label(scale(0.8)*"$Y$", (0,0), W); label(scale(0.8)*"$Z$", (10,0), E); label(scale(0.8)*"$A$", (2.5,2.2), W); label(scale(0.8)*"$B$", (7.5,2.2), E); label(scale(0.8)*"$C$", (5,0), S); fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white); /* Edit for solution */ /* A--C */ draw((2.5,2)--(5,0)); /* A--[half of YC] */ draw((2.5,2)--(2.5,0)); label(scale(0.8)*"$D$",(2.5,0),S); /* B--C */ draw((7.5,2)--(5,0)); /* B--[half of CZ] */ draw((7.5,2)--(7.5,0)); label(scale(0.8)*"$E$",(7.5,0),S); /* drawing right angles */ draw((2.5,0.5)--(2,0.5)); draw((2,0.5)--(2,0)); draw((2.5,0.5)--(3,0.5)); draw((3,0.5)--(3,0)); draw((7.5,0.5)--(7,0.5)); draw((7,0.5)--(7,0)); draw((7.5,0.5)--(8,0.5)); draw((8,0.5)--(8,0)); draw((5.5,2.5)--(4.5,2.5)--(4.5,1.5)--(5.5,1.5)--cycle); label(scale(0.8)*"$F$", (5,2), NE); label(scale(1)*"All credits for original unedited asymptote for the problem go to whoever made the asymptote in the 'Problem' section.", (5,-1), S); [/asy] We draw the triangles as shown above. What seems like a half-square (or a corner of a square) indicates $90$ °. (This is not a random idea - the two original two triangles on the top inspires us to do so.) Note that $\overline{AD}$ is perpendicular to $\overline{YC}$ $\overline{BE}$ is perpendicular to $\overline{CZ}$ $\overline{AB}$ is perpendicular to $\overline{YZ}$ , so because of corresponding angles theorem, $\angle YAD=\angle AXF$ . We are also told $\overline{YA}=\overline{AX}$ . Both $\bigtriangleup \text{XAF}$ and $\bigtriangleup \text{AYD}$ have a $90$ ° angle. Because of AAS Congruency (Angle-Angle-Side; $90$ °, $\angle YAD$ , and side $\overline{YA}$ ), $\bigtriangleup \text{XAF}$ is congruent to $\bigtriangleup \text{AYD}$ . Because of symmetry, the same goes for $\bigtriangleup \text{XFB}$ and $\bigtriangleup \text{BEZ}$ $\bigtriangleup \text{YAD}$ is congruent to $\bigtriangleup \text{CAD}$ , which is congruent to $\bigtriangleup \text{ACF}$ because of symmetry. So there are $4$ congruent triangles on the left side. There is also $4$ congruent triangles on the right side because of symmetry. There are $8$ congruent triangles in all. We are told the area of the triangle $\bigtriangleup \text{XYZ}$ is 8. $8\div8=1$ , so each of the small congruent triangles is $1$ . The shaded area contains $3$ small triangles, so the area of the shaded section is $1\cdot3=\boxed{3}$

D

3

9f39ac719326722c3bd53710e817ef59

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_22

Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom, and sides. [asy] /* AMC8 2002 #22 Problem */ draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw((0,1)--(0.5,1.5)--(1.5,1.5)--(1,1)); draw((1,0)--(1.5,0.5)--(1.5,1.5)); draw((0.5,1.5)--(1,2)--(1.5,2)); draw((1.5,1.5)--(1.5,3.5)--(2,4)--(3,4)--(2.5,3.5)--(2.5,0.5)--(1.5,.5)); draw((1.5,3.5)--(2.5,3.5)); draw((1.5,1.5)--(3.5,1.5)--(3.5,2.5)--(1.5,2.5)); draw((3,4)--(3,3)--(2.5,2.5)); draw((3,3)--(4,3)--(4,2)--(3.5,1.5)); draw((4,3)--(3.5,2.5)); draw((2.5,.5)--(3,1)--(3,1.5));[/asy] $\textbf{(A)}\ 18\qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 36$

Count the number of sides that are not exposed, where a cube is connected to another cube and subtract it from the total number of faces. There are $5$ places with two adjacent cubes, covering $10$ sides, and $(6)(6)=36$ faces. The exposed surface area is $36-10 = \boxed{26}$

C

26

9f39ac719326722c3bd53710e817ef59

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_22

Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom, and sides. [asy] /* AMC8 2002 #22 Problem */ draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw((0,1)--(0.5,1.5)--(1.5,1.5)--(1,1)); draw((1,0)--(1.5,0.5)--(1.5,1.5)); draw((0.5,1.5)--(1,2)--(1.5,2)); draw((1.5,1.5)--(1.5,3.5)--(2,4)--(3,4)--(2.5,3.5)--(2.5,0.5)--(1.5,.5)); draw((1.5,3.5)--(2.5,3.5)); draw((1.5,1.5)--(3.5,1.5)--(3.5,2.5)--(1.5,2.5)); draw((3,4)--(3,3)--(2.5,2.5)); draw((3,3)--(4,3)--(4,2)--(3.5,1.5)); draw((4,3)--(3.5,2.5)); draw((2.5,.5)--(3,1)--(3,1.5));[/asy] $\textbf{(A)}\ 18\qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 36$

We can count the number of showing faces from each side. One thing that we notice is that the front face has the same number of squares as the back face, the side faces have the same surface area, etc. Therefore, we are looking for $2($ front surface area $+$ side surface area $+$ top surface area $)$ . We find that this is $2(5 + 4 + 4) = 2 * 13 = \boxed{26}$

C

26

e694c12f846a101db0e11df782248a36

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_23

A corner of a tiled floor is shown. If the entire floor is tiled in this way and each of the four corners looks like this one, then what fraction of the tiled floor is made of darker tiles? [asy] /* AMC8 2002 #23 Problem */ fill((0,2)--(1,3)--(2,3)--(2,4)--(3,5)--(4,4)--(4,3)--(5,3)--(6,2)--(5,1)--(4,1)--(4,0)--(2,0)--(2,1)--(1,1)--cycle, mediumgrey); fill((7,1)--(6,2)--(7,3)--(8,3)--(8,4)--(9,5)--(10,4)--(7,0)--cycle, mediumgrey); fill((3,5)--(2,6)--(2,7)--(1,7)--(0,8)--(1,9)--(2,9)--(2,10)--(3,11)--(4,10)--(4,9)--(5,9)--(6,8)--(5,7)--(4,7)--(4,6)--cycle, mediumgrey); fill((6,8)--(7,9)--(8,9)--(8,10)--(9,11)--(10,10)--(10,9)--(11,9)--(11,7)--(10,7)--(10,6)--(9,5)--(8,6)--(8,7)--(7,7)--cycle, mediumgrey); draw((0,0)--(0,11)--(11,11)); for ( int x = 1; x &lt; 11; ++x ) { draw((x,11)--(x,0), linetype("4 4")); } for ( int y = 1; y &lt; 11; ++y ) { draw((0,y)--(11,y), linetype("4 4")); } clip((0,0)--(0,11)--(11,11)--(11,5)--(4,1)--cycle);[/asy] $\textbf{(A)}\ \frac{1}3\qquad\textbf{(B)}\ \frac{4}9\qquad\textbf{(C)}\ \frac{1}2\qquad\textbf{(D)}\ \frac{5}9\qquad\textbf{(E)}\ \frac{5}8$

The same pattern is repeated for every $6 \times 6$ tile. Looking closer, there is also symmetry of the top $3 \times 3$ square, so the fraction of the entire floor in dark tiles is the same as the fraction in the square. Counting the tiles, there are $4$ dark tiles, and $9$ total tiles, giving a fraction of $\boxed{49}$

B

49

7ac6ba05b6a53f53f296906619287770

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_24

Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract 8 ounces of pear juice from 3 pears and 8 ounces of orange juice from 2 oranges. She makes a pear-orange juice blend from an equal number of pears and oranges. What percent of the blend is pear juice? $\text{(A)}\ 30\qquad\text{(B)}\ 40\qquad\text{(C)}\ 50\qquad\text{(D)}\ 60\qquad\text{(E)}\ 70$

A pear gives $8/3$ ounces of juice per pear. An orange gives $8/2=4$ ounces of juice per orange. If the pear-orange juice blend used one pear and one orange each, the percentage of pear juice would be \[\frac{8/3}{8/3+4} \times 100 = \frac{8}{8+12} \times 100 = \boxed{40}\]

B

40

7ac6ba05b6a53f53f296906619287770

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_24

Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract 8 ounces of pear juice from 3 pears and 8 ounces of orange juice from 2 oranges. She makes a pear-orange juice blend from an equal number of pears and oranges. What percent of the blend is pear juice? $\text{(A)}\ 30\qquad\text{(B)}\ 40\qquad\text{(C)}\ 50\qquad\text{(D)}\ 60\qquad\text{(E)}\ 70$

Since it doesn't matter how many pears and oranges there are, you can make the number of them whatever you like. In this case, we could use $6$ , because it's the LCM of $2$ and $3$ . Then for the $6$ pears, there are $6/3*8=16$ ounces of pear juice. For the 6 oranges, there are $6/2*8=24$ ounces of orange juice. Since we are looking for the percent of pear juice, we need to do $16/(16+24)=16/40$ . Simplifying, we get $2/5$ . Hence the answer is $\boxed{40}$

B

40

e4725af41669e8d6a2780b4ad9991c89

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_25

Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have? $\text{(A)}\ \frac{1}{10}\qquad\text{(B)}\ \frac{1}{4}\qquad\text{(C)}\ \frac{1}{3}\qquad\text{(D)}\ \frac{2}{5}\qquad\text{(E)}\ \frac{1}{2}$

Since Ott gets equal amounts of money from each friend, we can say that he gets $x$ dollars from each friend. This means that Moe has $5x$ dollars, Loki has $4x$ dollars, and Nick has $3x$ dollars. The total amount is $12x$ dollars, and since Ott gets $3x$ dollars total, $\frac{3x}{12x}= \frac{3}{12} = \boxed{14}$

B

14

e4725af41669e8d6a2780b4ad9991c89

https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_25

Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have? $\text{(A)}\ \frac{1}{10}\qquad\text{(B)}\ \frac{1}{4}\qquad\text{(C)}\ \frac{1}{3}\qquad\text{(D)}\ \frac{2}{5}\qquad\text{(E)}\ \frac{1}{2}$

Assume Moe, Loki, and Nick each give Ott $$ 1$ . Therefore, Moe has $$ 5$ , Loki has $$ 4$ , and Nick has $$ 3$ . After everyone gives Ott some fraction of their money, the total money at the end situation will be the same as the original; which is $$ 12$ . Ott gets $$ 1$ $+$ $$ 1$ $+$ $$ 1$ $=$ $$ 3$ . Thus, the answer is $\frac{3}{12}=\boxed{14}$

B

14

466eda3a9acd6cf77c28bed6f9f08845

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_1

Casey's shop class is making a golf trophy. He has to paint 300 dimples on a golf ball. If it takes him 2 seconds to paint one dimple, how many minutes will he need to do his job? $\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 12 }$

It will take him $300\cdot2=600$ seconds to paint all the dimples. This is equivalent to $\frac{600}{60}=10$ minutes $\Rightarrow \boxed{10}$

D

10

774ea68b4309ab5ddd17e91218b14bcc

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_2

I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number? $\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12$

Let the numbers be $x$ and $y$ . Then we have $x+y=11$ and $xy=24$ . Solving for $x$ in the first equation yields $x=11-y$ , and substituting this into the second equation gives $(11-y)(y)=24$ . Simplifying this gives $-y^2+11y=24$ , or $y^2-11y+24=0$ . This factors as $(y-3)(y-8)=0$ , so $y=3$ or $y=8$ , and the corresponding $x$ values are $x=8$ and $x=3$ . These are essentially the same answer: one number is $3$ and one number is $8$ , so the largest number is $8, \boxed{8}$

D

8

774ea68b4309ab5ddd17e91218b14bcc

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_2

I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number? $\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12$

Use the answers to attempt to "reverse engineer" an appropriate pair of numbers. Looking at option $A$ , guess that one of the numbers is $3$ . If the sum of two numbers is $11$ and one is $3$ , then other must be $11 - 3 = 8$ . The product of those numbers is $3\cdot 8 = 24$ , which is the second condition of the problem, so our number are $3$ and $8$ However, $3$ is the smaller of the two numbers, so the answer is $8$ or $\boxed{8}$

D

8

0810cde0d4b414eae8f5a1c27937971b

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_3

Granny Smith has $63. Elberta has $2 more than Anjou and Anjou has one-third as much as Granny Smith. How many dollars does Elberta have? $\text{(A)}\ 17 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 19 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 23$

Since Anjou has $\frac{1}{3}$ the amount of money as Granny Smith and Granny Smith has $ $63$ , Anjou has $\frac{1}{3}\times63=21$ dollars. Elberta has $ $2$ more than this, so she has $ $23$ , or $\boxed{23}$

E

23

24d17ae7129a7a8139295df59d58a0df

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_4

The digits 1, 2, 3, 4 and 9 are each used once to form the smallest possible even five-digit number. The digit in the tens place is $\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 9$

Since the number is even, the last digit must be $2$ or $4$ . To make the smallest possible number, the ten-thousands digit must be as small as possible, so the ten-thousands digit is $1$ . Simillarly, the thousands digit has second priority, so it must also be as small as possible once the ten-thousands digit is decided, so the thousands digit is $2$ . Similarly, the hundreds digit needs to be the next smallest number, so it is $3$ . However, for the tens digit, we can't use $4$ , since we already used $2$ and the number must be even, so the units digit must be $4$ and the tens digit is $9, \boxed{9}$ (The number is $12394$ .)

E

9

a1f117c233009d82b173a6e1303475c3

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_6

Six trees are equally spaced along one side of a straight road. The distance from the first tree to the fourth is 60 feet. What is the distance in feet between the first and last trees? $\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 105 \qquad \text{(D)}\ 120 \qquad \text{(E)}\ 140$

There are $3$ spaces between the 1st and 4th trees, so each of these spaces has $\frac{60}{3}=20$ feet. Between the first and last trees there are $5$ spaces, so the distance between them is $20\times5=100$ feet, $\boxed{100}$

B

100

60811600c81436d8b436a59165b7c6ad

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_7

To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid. [asy] for (int a = 0; a < 7; ++a) { for (int b = 0; b < 8; ++b) { dot((a,b)); } } draw((3,0)--(0,5)--(3,7)--(6,5)--cycle); [/asy] What is the number of square inches in the area of the small kite? $\text{(A)}\ 21 \qquad \text{(B)}\ 22 \qquad \text{(C)}\ 23 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 25$

The area of a kite is half the product of its diagonals. The diagonals have lengths of $6$ and $7$ , so the area is $\frac{(6)(7)}{2}=21, \boxed{21}$

A

21

60811600c81436d8b436a59165b7c6ad

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_7

To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid. [asy] for (int a = 0; a < 7; ++a) { for (int b = 0; b < 8; ++b) { dot((a,b)); } } draw((3,0)--(0,5)--(3,7)--(6,5)--cycle); [/asy] What is the number of square inches in the area of the small kite? $\text{(A)}\ 21 \qquad \text{(B)}\ 22 \qquad \text{(C)}\ 23 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 25$

Drawing in the diagonals of the kite will form four right triangles on the "inside" part of the grid. Drawing in the border of the 7 by 6 grid will form four right triangles on the "outside" part of the grid. Since each right triangle on the inside can be paired with a congruent right triangle that is on the outside, the area of the kite is half the total area of the grid, or $\frac{(6)(7)}{2}=21, \boxed{21}$

A

21

60811600c81436d8b436a59165b7c6ad

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_7

To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid. [asy] for (int a = 0; a < 7; ++a) { for (int b = 0; b < 8; ++b) { dot((a,b)); } } draw((3,0)--(0,5)--(3,7)--(6,5)--cycle); [/asy] What is the number of square inches in the area of the small kite? $\text{(A)}\ 21 \qquad \text{(B)}\ 22 \qquad \text{(C)}\ 23 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 25$

Pick's Theorem states: \[\frac{\text{number of boundary points}}{2}+\text{number of interior points}-1\] as the area of a figure on a grid. Counting, we see there are $4$ boundary points and $20$ interior points. Therefore, we have \[\frac{4}{2}+20-1\implies 20+1\implies 21.\] Hence, the answer is $\boxed{21}$

A

21

5cb202a7ae468acc61c370a9559e49dc

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_8

To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid. [asy] size(85); for (int a = 0; a < 7; ++a) { for (int b = 0; b < 8; ++b) { dot((a,b)); } } draw((3,0)--(0,5)--(3,7)--(6,5)--cycle); [/asy] Genevieve puts bracing on her large kite in the form of a cross connecting opposite corners of the kite. How many inches of bracing material does she need? $\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 38 \qquad \text{(E)}\ 39$

Each diagonal of the large kite is $3$ times the length of the corresponding diagonal of the short kite since it was made with a grid $3$ times as long in each direction. The diagonals of the small kite are $6$ and $7$ , so the diagonals of the large kite are $18$ and $21$ , and the amount of bracing Genevieve needs is the sum of these lengths, which is $39, \boxed{39}$

E

39

39ca63cab5fbe12c72438d8baa973c15

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_9

To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid. [asy] for (int a = 0; a < 7; ++a) { for (int b = 0; b < 8; ++b) { dot((a,b)); } } draw((3,0)--(0,5)--(3,7)--(6,5)--cycle); [/asy] The large kite is covered with gold foil. The foil is cut from a rectangular piece that just covers the entire grid. How many square inches of waste material are cut off from the four corners? $\textbf{(A)}\ 63 \qquad \textbf{(B)}\ 72 \qquad \textbf{(C)}\ 180 \qquad \textbf{(D)}\ 189 \qquad \textbf{(E)}\ 264$

The large grid has dimensions three times that of the small grid, so its dimensions are $3(6)\times3(7)$ , or $18\times21$ , so the area is $(18)(21)=378$ . The area of the kite is half of the area of the rectangle as you can see, so the area of the waste material is also half the area of the rectangle. Thus, the area of the waste material is $378/2=\boxed{189}$

D

189

d121538571fa5ab9ec81a162a70ad694

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_10

A collector offers to buy state quarters for 2000% of their face value. At that rate how much will Bryden get for his four state quarters? $\text{(A)}\ 20\text{ dollars} \qquad \text{(B)}\ 50\text{ dollars} \qquad \text{(C)}\ 200\text{ dollars} \qquad \text{(D)}\ 500\text{ dollars} \qquad \text{(E)}\ 2000\text{ dollars}$

$2000\%$ is equivalent to $20\times100\%$ . Therefore, $2000\%$ of a number is the same as $20$ times that number. $4$ quarters is $1$ dollar, so Bryden will get $20\times1={20}$ dollars, $\boxed{20}$

A

20

d121538571fa5ab9ec81a162a70ad694

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_10

A collector offers to buy state quarters for 2000% of their face value. At that rate how much will Bryden get for his four state quarters? $\text{(A)}\ 20\text{ dollars} \qquad \text{(B)}\ 50\text{ dollars} \qquad \text{(C)}\ 200\text{ dollars} \qquad \text{(D)}\ 500\text{ dollars} \qquad \text{(E)}\ 2000\text{ dollars}$

Since $2000\%$ is just $\frac{2000}{100}$ , we can multiply that by $100$ , because four quarters is a $100$ cents. After the multiplication, we get $2000$ . Since our answer is in cents right now, we need to convert it to dollars, which would be $\boxed{20}$ dollars.

A

20

5b9156da5bbdd897db4341bdb83b8f44

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_11

Points $A$ $B$ $C$ and $D$ have these coordinates: $A(3,2)$ $B(3,-2)$ $C(-3,-2)$ and $D(-3, 0)$ . The area of quadrilateral $ABCD$ is [asy] for (int i = -4; i <= 4; ++i) { for (int j = -4; j <= 4; ++j) { dot((i,j)); } } draw((0,-4)--(0,4),linewidth(1)); draw((-4,0)--(4,0),linewidth(1)); for (int i = -4; i <= 4; ++i) { draw((i,-1/3)--(i,1/3),linewidth(0.5)); draw((-1/3,i)--(1/3,i),linewidth(0.5)); } [/asy] $\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 24$

[asy] for (int i = -4; i <= 4; ++i) { for (int j = -4; j <= 4; ++j) { dot((i,j)); } } draw((0,-4)--(0,4),linewidth(1)); draw((-4,0)--(4,0),linewidth(1)); for (int i = -4; i <= 4; ++i) { draw((i,-1/3)--(i,1/3),linewidth(0.5)); draw((-1/3,i)--(1/3,i),linewidth(0.5)); } { draw((3,2)--(3,-2)--(-3,-2)--(-3,0)--cycle,linewidth(1)); } label("$A$",(3,2),NE); label("$B$",(3,-2),SE); label("$C$",(-3,-2),SW); label("$D$",(-3,0),NW); [/asy] This quadrilateral is a trapezoid, because $AB\parallel CD$ but $BC$ is not parallel to $AD$ . The area of a trapezoid is the product of its height and its median, where the median is the average of the side lengths of the bases. The two bases are $AB$ and $CD$ , which have lengths $2$ and $4$ , respectively, so the length of the median is $\frac{2+4}{2}=3$ $CB$ is perpendicular to the bases, so it is the height, and has length $6$ . Therefore, the area of the trapezoid is $(3)(6)=18, \boxed{18}$

C

18

5b9156da5bbdd897db4341bdb83b8f44

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_11

Points $A$ $B$ $C$ and $D$ have these coordinates: $A(3,2)$ $B(3,-2)$ $C(-3,-2)$ and $D(-3, 0)$ . The area of quadrilateral $ABCD$ is [asy] for (int i = -4; i <= 4; ++i) { for (int j = -4; j <= 4; ++j) { dot((i,j)); } } draw((0,-4)--(0,4),linewidth(1)); draw((-4,0)--(4,0),linewidth(1)); for (int i = -4; i <= 4; ++i) { draw((i,-1/3)--(i,1/3),linewidth(0.5)); draw((-1/3,i)--(1/3,i),linewidth(0.5)); } [/asy] $\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 24$

Using the diagram above, the figure can be divided along the x-axis into two familiar regions that do not overlap: a right triangle and a rectangle. Since the areas do not overlap, the area of the entire trapezoid is the sum of the area of the triangle and the area of the rectangle. $A_{trap} = A_{tri} + A_{rect}$ $A_{trap} = \frac{1}{2}bh + lw$ $A_{trap} = \frac{1}{2}\cdot 6 \cdot 2 + 6\cdot 2$ $A_{trap} = 6 + 12 = 18 \rightarrow \boxed{18}$

C

18

cc3ede93eb388d1dda4886d1ddb66085

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_12

If $a\otimes b = \dfrac{a + b}{a - b}$ , then $(6\otimes 4)\otimes 3 =$ $\text{(A)}\ 4 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 72$

$6\otimes4=\frac{6+4}{6-4}=5$ $5\otimes3=\frac{5+3}{5-3}=4, \boxed{4}$

A

4

904b5925faf06c5d9be7a6098ebb10a3

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_13

Of the 36 students in Richelle's class, 12 prefer chocolate pie, 8 prefer apple, and 6 prefer blueberry. Half of the remaining students prefer cherry pie and half prefer lemon. For Richelle's pie graph showing this data, how many degrees should she use for cherry pie? $\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 50 \qquad \text{(E)}\ 72$

There are $36$ students in the class: $12$ prefer chocolate pie, $8$ prefer apple pie, and $6$ prefer blueberry pie. Therefore, $36-12-8-6=10$ students prefer cherry pie or lemon pie. Half of these prefer each, so $5$ students prefer cherry pie. This means that $\frac{5}{36}$ of the students prefer cherry pie, so $\frac{5}{36}$ of the full $360^\circ$ should be used for cherry pie. This is $(\frac{5}{36})(360^\circ)=50^\circ, \boxed{50}$

D

50

56f3dfc4a77a4a538b7fea152fdd17bf

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_14

Tyler has entered a buffet line in which he chooses one kind of meat, two different vegetables and one dessert. If the order of food items is not important, how many different meals might he choose? $\text{(A)}\ 4 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 72 \qquad \text{(D)}\ 80 \qquad \text{(E)}\ 144$

There are $3$ possibilities for the meat and $4$ possibilites for the dessert, for a total of $4\times3=12$ possibilities for the meat and the dessert. There are $4$ possibilities for the first vegetable and $3$ possibilities for the second, but order doesn't matter, so we overcounted by a factor of $2$ . For example, we counted 'baked beans and corn' and 'corn and baked beans' as $2$ different possibilities, so the total possibilites for the two vegetables is $\frac{4\times3}{2}=6$ , and the total number of possibilites is $12\times6=\boxed{72}.$

C

72

79d50274a631563b346307994de06ac3

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_15

Homer began peeling a pile of 44 potatoes at the rate of 3 potatoes per minute. Four minutes later Christen joined him and peeled at the rate of 5 potatoes per minute. When they finished, how many potatoes had Christen peeled? $\text{(A)}\ 20 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 33 \qquad \text{(E)}\ 40$

After the $4$ minutes of Homer peeling alone, he had peeled $4\times3=12$ potatoes. This means that there are $44-12=32$ potatoes left. Once Christen joins him, the two are peeling potatoes at a rate of $3+5=8$ potatoes per minute. So, they finish peeling after another $\frac{32}{8}=4$ minutes. In these $4$ minutes, Christen peeled $4\times5=20$ potatoes, $\boxed{20}$

A

20

7bfd03b2ce4f8c5336bc7b345b0e32ea

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_21

The mean of a set of five different positive integers is 15. The median is 18. The maximum possible value of the largest of these five integers is $\text{(A)}\ 19 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 40$

Since there is an odd number of terms, the median is the number in the middle, specifically, the third largest number is $18$ , and there are $2$ numbers less than $18$ and $2$ numbers greater than $18$ . The sum of these integers is $5(15)=75$ , since the mean is $15$ . To make the largest possible number with a given sum, the other numbers must be as small as possible. The two numbers less than $18$ must be positive and distinct, so the smallest possible numbers for these are $1$ and $2$ . The number right after $18$ also needs to be as small as possible, so it must be $19$ . This means that the remaining number, the maximum possible value for a number in the set, is $75-1-2-18-19=35, \boxed{35}$

D

35

c785c37af78d511634c62cfc488dca61

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_22

On a twenty-question test, each correct answer is worth 5 points, each unanswered question is worth 1 point and each incorrect answer is worth 0 points. Which of the following scores is NOT possible? $\text{(A)}\ 90 \qquad \text{(B)}\ 91 \qquad \text{(C)}\ 92 \qquad \text{(D)}\ 95 \qquad \text{(E)}\ 97$

The highest possible score is if you get every answer right, to get $5(20)=100$ . The second highest possible score is if you get $19$ questions right and leave the remaining one blank, to get a $5(19)+1(1)=96$ . Therefore, no score between $96$ and $100$ , exclusive, is possible, so $97$ is not possible, $\boxed{97}$

E

97

c785c37af78d511634c62cfc488dca61

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_22

On a twenty-question test, each correct answer is worth 5 points, each unanswered question is worth 1 point and each incorrect answer is worth 0 points. Which of the following scores is NOT possible? $\text{(A)}\ 90 \qquad \text{(B)}\ 91 \qquad \text{(C)}\ 92 \qquad \text{(D)}\ 95 \qquad \text{(E)}\ 97$

We can equivalently construct the following rules: You have 100 point at first, but if you give the wrong answer, you will lose 5 points, if you don't answer a question you will lose 4 points. Obviously, you can lose 10 points, 9 points, 8 points, 5 points or 4 points, but you cannot lose 3 points. The answer is $\boxed{97}$

E

97

9097088d5eb19609439e4b7723d3f244

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_23

Points $R$ $S$ and $T$ are vertices of an equilateral triangle, and points $X$ $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be drawn using any three of these six points as vertices? [asy] pair SS,R,T,X,Y,Z; SS = (2,2*sqrt(3)); R = (0,0); T = (4,0); X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(3)); dot(SS); dot(R); dot(T); dot(X); dot(Y); dot(Z); label("$S$",SS,N); label("$R$",R,SW); label("$T$",T,SE); label("$X$",X,S); label("$Y$",Y,NW); label("$Z$",Z,NE); [/asy] $\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20$

There are $6$ points in the figure, and $3$ of them are needed to form a triangle, so there are ${6\choose{3}} =20$ possible triplets of the $6$ points. However, some of these created congruent triangles, and some don't even make triangles at all. Case 1: Triangles congruent to $\triangle RST$ There is obviously only $1$ of these: $\triangle RST$ itself. Case 2: Triangles congruent to $\triangle SYZ$ There are $4$ of these: $\triangle SYZ, \triangle RXY, \triangle TXZ,$ and $\triangle XYZ$ Case 3: Triangles congruent to $\triangle RSX$ There are $6$ of these: $\triangle RSX, \triangle TSX, \triangle STY, \triangle RTY, \triangle RSZ,$ and $\triangle RTZ$ Case 4: Triangles congruent to $\triangle SYX$ There are again $6$ of these: $\triangle SYX, \triangle SZX, \triangle TYZ, \triangle TYX, \triangle RXZ,$ and $\triangle RYZ$ However, if we add these up, we accounted for only $1+4+6+6=17$ of the $20$ possible triplets. We see that the remaining triplets don't even form triangles; they are $SYR, RXT,$ and $TZS$ . Adding these $3$ into the total yields for all of the possible triplets, so we see that there are only $4$ possible non-congruent, non-degenerate triangles, $\boxed{4}$

D

4

9097088d5eb19609439e4b7723d3f244

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_23

Points $R$ $S$ and $T$ are vertices of an equilateral triangle, and points $X$ $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be drawn using any three of these six points as vertices? [asy] pair SS,R,T,X,Y,Z; SS = (2,2*sqrt(3)); R = (0,0); T = (4,0); X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(3)); dot(SS); dot(R); dot(T); dot(X); dot(Y); dot(Z); label("$S$",SS,N); label("$R$",R,SW); label("$T$",T,SE); label("$X$",X,S); label("$Y$",Y,NW); label("$Z$",Z,NE); [/asy] $\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20$

We can do casework in this problem like the solution above, but that would take too much time. Instead, we see that we can split this equilateral dot triangle to two halves by dropping an altitude from the top vertex of the big triangle. Using the smaller triangle we won't have to worry about extra unneeded cases. We can see that there are three distinct triangles in the half, and combining this with the larger equilateral triangle our answer is $3+1 = \boxed{4}$

D

4

9097088d5eb19609439e4b7723d3f244

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_23

Points $R$ $S$ and $T$ are vertices of an equilateral triangle, and points $X$ $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be drawn using any three of these six points as vertices? [asy] pair SS,R,T,X,Y,Z; SS = (2,2*sqrt(3)); R = (0,0); T = (4,0); X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(3)); dot(SS); dot(R); dot(T); dot(X); dot(Y); dot(Z); label("$S$",SS,N); label("$R$",R,SW); label("$T$",T,SE); label("$X$",X,S); label("$Y$",Y,NW); label("$Z$",Z,NE); [/asy] $\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20$

Notice that 20 is obviously too high (There are only 20 ways to choose 3 of the points to form a triangle or a line in total!) and you can count 4 distinct triangles quickly: $\triangle RYX$ $\triangle RYT$ $\triangle RYZ$ $\triangle RST$ . So the answer is $\boxed{4}$

D

4

f723f301d3b11e3f7dc13a5fb31af0d4

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_24

Each half of this figure is composed of 3 red triangles, 5 blue triangles and 8 white triangles. When the upper half is folded down over the centerline, 2 pairs of red triangles coincide, as do 3 pairs of blue triangles. There are 2 red-white pairs. How many white pairs coincide? [asy] draw((0,0)--(4,4*sqrt(3))); draw((1,-sqrt(3))--(5,3*sqrt(3))); draw((2,-2*sqrt(3))--(6,2*sqrt(3))); draw((3,-3*sqrt(3))--(7,sqrt(3))); draw((4,-4*sqrt(3))--(8,0)); draw((8,0)--(4,4*sqrt(3))); draw((7,-sqrt(3))--(3,3*sqrt(3))); draw((6,-2*sqrt(3))--(2,2*sqrt(3))); draw((5,-3*sqrt(3))--(1,sqrt(3))); draw((4,-4*sqrt(3))--(0,0)); draw((3,3*sqrt(3))--(5,3*sqrt(3))); draw((2,2*sqrt(3))--(6,2*sqrt(3))); draw((1,sqrt(3))--(7,sqrt(3))); draw((-1,0)--(9,0)); draw((1,-sqrt(3))--(7,-sqrt(3))); draw((2,-2*sqrt(3))--(6,-2*sqrt(3))); draw((3,-3*sqrt(3))--(5,-3*sqrt(3))); [/asy] $\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 9$

Each half has $3$ red triangles, $5$ blue triangles, and $8$ white triangles. There are also $2$ pairs of red triangles, so $2$ red triangles on each side are used, leaving $1$ red triangle, $5$ blue triangles, and $8$ white triangles remaining on each half. Also, there are $3$ pairs of blue triangles, using $3$ blue triangles on each side, so there is $1$ red triangle, $2$ blue triangles, and $8$ white triangles remaining on each half. Also, we have $2$ red-white pairs. This obviously can't use $2$ red triangles on one side, since there is only $1$ on each side, so we must use $1$ red triangle and $1$ white triangle per side, leaving $2$ blue triangles and $7$ white triangles apiece. The remaining blue triangles cannot be matched with other blue triangles since that would mean there were more than $3$ blue pairs, so the remaining blue triangles must be paired with white triangles, yielding $4$ blue-white pairs, one for each of the remaining blue triangles. This uses $2$ blue triangles and $2$ white triangles on each side, leaving $5$ white triangles apiece, which must be paired with each other, so there are $5$ white-white pairs, $\boxed{5}$

B

5

f6886b12e08ebe1b17b506a8c487051f

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_25

There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it? $\textbf{(A)}\ 5724 \qquad \textbf{(B)}\ 7245 \qquad \textbf{(C)}\ 7254 \qquad \textbf{(D)}\ 7425 \qquad \textbf{(E)}\ 7542$

We begin by narrowing down the possibilities. If the larger number were twice the smaller number, then the smallest possibility for the larger number is $2457\times2=4914$ , since $2457$ is the smallest number in the set. The largest possibility would have to be twice the largest number in the set such that when it is multiplied by $2$ , it is less than or equal to $7542$ , the largest number in the set. This happens to be $2754\times2=5508$ . Therefore, the number would have to be between $4914$ and $5508$ , and also even. The only even numbers in the set and in this range are $5472$ and $5274$ . A quick check reveals that neither of these numbers is twice a number in the set. The number can't be quadruple or more another number in the set since $2457\times4=9828$ , well past the range of the set. Therefore, the number must be triple another number in the set. The least possibility is $2457\times3=7371$ and the greatest is $2475\times3=7425$ , since any higher number in the set multiplied by $3$ would be out of the range of the set. Reviewing, we find that the upper bound does in fact work, so the multiple is $2475\times3=7425, \boxed{7425}$

D

7425

f6886b12e08ebe1b17b506a8c487051f

https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_25

There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it? $\textbf{(A)}\ 5724 \qquad \textbf{(B)}\ 7245 \qquad \textbf{(C)}\ 7254 \qquad \textbf{(D)}\ 7425 \qquad \textbf{(E)}\ 7542$

There are only 5 options for the problem so we can just try them. It is easy since that we only need try to use $2$ $3$ to divide them. Even $5$ will leads to a solution start with $1$ which we don't need. $5724/2=2862$ $5724/3=1908$ $7245/3=2415$ $7254/2=3612$ $7254/3=2418$ $7425/3=2475$ . The answer is $\boxed{7425}$ . You can obtain the answer in only 6 calculations.

D

7425

d7fc7b0dabdd8b7a14413e8ad286fee0

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_1

Aunt Anna is $42$ years old. Caitlin is $5$ years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin? $\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 16\qquad\mathrm{(C)}\ 17\qquad\mathrm{(D)}\ 21\qquad\mathrm{(E)}\ 37$

If Brianna is half as old as Aunt Anna, then Brianna is $\frac{42}{2}$ years old, or $21$ years old. If Caitlin is $5$ years younger than Brianna, she is $21-5$ years old, or $16$ So, the answer is $\boxed{16}$

B

16

d7fc7b0dabdd8b7a14413e8ad286fee0

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_1

Aunt Anna is $42$ years old. Caitlin is $5$ years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin? $\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 16\qquad\mathrm{(C)}\ 17\qquad\mathrm{(D)}\ 21\qquad\mathrm{(E)}\ 37$

Since Brianna is half of Aunt Anna's age this means that Brianna is $21$ years old. Now we just find Caitlin's age by doing $21-5$ . This makes $16$ or $\boxed{16}$

B

16

c228c1b86653c3d32d090fa94df348be

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_2

Which of these numbers is less than its reciprocal? $\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

The number $0$ has no reciprocal, and $1$ and $-1$ are their own reciprocals. This leaves only $2$ and $-2$ . The reciprocal of $2$ is $1/2$ , but $2$ is not less than $1/2$ . The reciprocal of $-2$ is $-1/2$ , and $-2$ is less than $-1/2$ , so it is $\boxed{2}$

A

2

c228c1b86653c3d32d090fa94df348be

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_2

Which of these numbers is less than its reciprocal? $\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

The statement "a number is less than its reciprocal" can be translated as $x < \frac{1}{x}$ Multiplication by $x$ can be done if you do it in three parts: $x>0$ $x=0$ , and $x<0$ . You have to be careful about the direction of the inequality, as you do not know the sign of $x$ If $x>0$ , the sign of the inequality remains the same. Thus, we have $x^2 < 1$ when $x > 0$ . This leads to $0 < x < 1$ If $x=0$ , the inequality $x < \frac{1}{x}$ is undefined. If $x<0$ , the sign of the inequality must be switched. Thus, we have $x^2 > 1$ when $x < 0$ . This leads to $x < -1$ Putting the solutions together, we have $x<-1$ or $0 < x < 1$ , or in interval notation, $(-\infty, -1) \cup(0, 1)$ . The only answer in that range is $\boxed{2}$

A

2

c228c1b86653c3d32d090fa94df348be

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_2

Which of these numbers is less than its reciprocal? $\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

Starting again with $x < \frac{1}{x}$ , we avoid multiplication by $x$ . Instead, move everything to the left, and find a common denominator: $x < \frac{1}{x}$ $x - \frac{1}{x} < 0$ $\frac{x^2 - 1}{x} < 0$ $\frac{(x+1)(x-1)}{x} < 0$ Divide this expression at $x=-1$ $x=0$ , and $x=1$ , as those are the three points where the expression on the left will "change sign". If $x<-1$ , all three of those terms will be negative, and the inequality is true. Therefore, $(-\infty, -1)$ is part of our solution set. If $-1 < x < 0$ , the $(x+1)$ term will become positive, but the other two terms remain negative. Thus, there are no solutions in this region. If $0 < x < 1$ , then both $(x+1)$ and $x$ are positive, while $(x-1)$ remains negative. Thus, the entire region $(0, -1)$ is part of the solution set. If $1 < x$ , then all three terms are positive, and there are no solutions. At all three "boundary points", the function is either $0$ or undefined. Therefore, the entire solution set is $(-\infty, -1) \cup (0, 1)$ , and the only option in that region is $x=-2$ , leading to $\boxed{2}$

A

2

c228c1b86653c3d32d090fa94df348be

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_2

Which of these numbers is less than its reciprocal? $\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

We can find out all of their reciprocals. Now we compare and see that the answer is $\boxed{2}$

A

2

e939f57d8e4494d93d0becb9af4651a5

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_3

How many whole numbers lie in the interval between $\frac{5}{3}$ and $2\pi$ $\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ \text{infinitely many}$

The smallest whole number in the interval is $2$ because $5/3$ is more than $1$ but less than $2$ . The largest whole number in the interval is $6$ because $2\pi$ is more than $6$ but less than $7$ . There are five whole numbers in the interval. They are $2$ $3$ $4$ $5$ , and $6$ , so the answer is $\boxed{5}$

D

5

98cfee3e7389abeb56f9bb2197c812e7

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_5

Each principal of Lincoln High School serves exactly one $3$ -year term. What is the maximum number of principals this school could have during an $8$ -year period? $\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 8$

If the first year of the $8$ -year period was the final year of a principal's term, then in the next six years two more principals would serve, and the last year of the period would be the first year of the fourth principal's term. Therefore, the maximum number of principals who can serve during an $8$ -year period is $4$ , so the answer is $\boxed{4}$ if the terms are divided $1\ |\ 2\ 3\ 4\ |\ 5\ 6\ 7\ |\ 8$

C

4

0c1e231529af9f2fcf9875935cc774f1

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_6

Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded $L$ -shaped region is [asy] pair A,B,C,D; A = (5,5); B = (5,0); C = (0,0); D = (0,5); fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); draw(A--B--C--D--cycle); draw((4,0)--(4,4)--(0,4)); draw((1,5)--(1,1)--(5,1)); label("$A$",A,NE); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,NW); label("$1$",(1,4.5),E); label("$1$",(0.5,5),N); label("$3$",(1,2.5),E); label("$3$",(2.5,1),N); label("$1$",(4,0.5),E); label("$1$",(4.5,1),N); [/asy] $\text{(A)}\ 7 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

The side of the large square is $1 + 3 + 1 = 5$ , so the area of the large square is $5^2 = 25$ The area of the middle square is $3^2$ , and the sum of the areas of the two smaller squares is $2 * 1^2 = 2$ Thus, the big square minus the three smaller squares is $25 - 9 - 2 = 14$ . This is the area of the two congruent L-shaped regions. So the area of one L-shaped region is $\frac{14}{2} = 7$ , and the answer is $\boxed{7}$

A

7

0c1e231529af9f2fcf9875935cc774f1

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_6

Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded $L$ -shaped region is [asy] pair A,B,C,D; A = (5,5); B = (5,0); C = (0,0); D = (0,5); fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); draw(A--B--C--D--cycle); draw((4,0)--(4,4)--(0,4)); draw((1,5)--(1,1)--(5,1)); label("$A$",A,NE); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,NW); label("$1$",(1,4.5),E); label("$1$",(0.5,5),N); label("$3$",(1,2.5),E); label("$3$",(2.5,1),N); label("$1$",(4,0.5),E); label("$1$",(4.5,1),N); [/asy] $\text{(A)}\ 7 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

The shaded area can be divided into two regions: one rectangle that is 1 by 3, and one rectangle that is 4 by 1. (Or the reverse, depending on which rectangle the 1 by 1 square is "joined" to.) Either way, the total area of these two regions is $3 + 4 = 7$ , and the answer is $\boxed{7}$

A

7

0c1e231529af9f2fcf9875935cc774f1

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_6

Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded $L$ -shaped region is [asy] pair A,B,C,D; A = (5,5); B = (5,0); C = (0,0); D = (0,5); fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); draw(A--B--C--D--cycle); draw((4,0)--(4,4)--(0,4)); draw((1,5)--(1,1)--(5,1)); label("$A$",A,NE); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,NW); label("$1$",(1,4.5),E); label("$1$",(0.5,5),N); label("$3$",(1,2.5),E); label("$3$",(2.5,1),N); label("$1$",(4,0.5),E); label("$1$",(4.5,1),N); [/asy] $\text{(A)}\ 7 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

Chop the entire 5 by 5 region into $25$ squares like a piece of graph paper. When you draw all the lines, you can count that only $7$ of the small 1 by 1 squares will be shaded, giving $\boxed{7}$ as the answer.

A

7

0c1e231529af9f2fcf9875935cc774f1

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_6

Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded $L$ -shaped region is [asy] pair A,B,C,D; A = (5,5); B = (5,0); C = (0,0); D = (0,5); fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); draw(A--B--C--D--cycle); draw((4,0)--(4,4)--(0,4)); draw((1,5)--(1,1)--(5,1)); label("$A$",A,NE); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,NW); label("$1$",(1,4.5),E); label("$1$",(0.5,5),N); label("$3$",(1,2.5),E); label("$3$",(2.5,1),N); label("$1$",(4,0.5),E); label("$1$",(4.5,1),N); [/asy] $\text{(A)}\ 7 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

In the bottpom left corner of the 5 by 5 square there is a 4 by 4 square which has an area of $4\cdot4=16$ . In the top right of that 4 by 4 square is a 3 by 3 square with an area of $3\cdot3=9$ . When we remove the 3 by 3 square from the 4 by 4 square we get the L-shaped figure so our answer is $16-9=\boxed{7}$

A

7

b4161fa2595b9caf0fedb65417094192

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_7

What is the minimum possible product of three different numbers of the set $\{-8,-6,-4,0,3,5,7\}$ $\text{(A)}\ -336 \qquad \text{(B)}\ -280 \qquad \text{(C)}\ -210 \qquad \text{(D)}\ -192 \qquad \text{(E)}\ 0$

The only way to get a negative product using three numbers is to multiply one negative number and two positives or three negatives. Only two reasonable choices exist: $(-8)\times(-6)\times(-4) = (-8)\times(24) = -192$ and $(-8)\times5\times7 = (-8)\times35 = -280$ . The latter is smaller, so $\boxed{280}$

B

280

f530163b6184f697cd4b5054402e0ccf

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_8

problem_id f530163b6184f697cd4b5054402e0ccf Three dice with faces numbered $1$ through $6$... f530163b6184f697cd4b5054402e0ccf Three dice with faces numbered 1 through 6 are... Name: Text, dtype: object

The numbers on one die total $1+2+3+4+5+6 = 21$ , so the numbers on the three dice total $63$ . Numbers $1, 1, 2, 3, 4, 5, 6$ are visible, and these total $22$ . This leaves $63 - 22 = \boxed{41}$ not seen.

D

41

854a2ebf565fd977456d93a7504b0415

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_9

Three-digit powers of $2$ and $5$ are used in this "cross-number" puzzle. What is the only possible digit for the outlined square? \[\begin{array}{lcl} \textbf{ACROSS} & & \textbf{DOWN} \\ \textbf{2}.~ 2^m & & \textbf{1}.~ 5^n \end{array}\] [asy] draw((0,-1)--(1,-1)--(1,2)--(0,2)--cycle); draw((0,1)--(3,1)--(3,0)--(0,0)); draw((3,0)--(2,0)--(2,1)--(3,1)--cycle,linewidth(2)); label("$1$",(0,2),SE); label("$2$",(0,1),SE); [/asy] $\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

The $3$ -digit powers of $5$ are $125$ and $625$ , so space $2$ is filled with a $2$ . The only $3$ -digit power of $2$ beginning with $2$ is $256$ , so the outlined block is filled with a $\boxed{6}$

D

6

2a7e17e4bcec10c67eb73241e3c34073

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_10

Ara and Shea were once the same height. Since then Shea has grown 20% while Ara has grown half as many inches as Shea. Shea is now 60 inches tall. How tall, in inches, is Ara now? $\text{(A)}\ 48 \qquad \text{(B)}\ 51 \qquad \text{(C)}\ 52 \qquad \text{(D)}\ 54 \qquad \text{(E)}\ 55$

Shea has grown $20\%$ , if x was her original height, then $1.2x = 60$ , so she was originally $\frac{60}{1.2}=50$ inches tall which is a $60 - 50 = 10$ inch increase. Ara also started off at $50$ inches. Since Ara grew half as much as Shea, Ara grew $\frac{10}{2} = 5$ inches. Therefore, Ara is now $50+5=55$ inches tall which is choice $\boxed{55}.$

E

55

268a7294c8e8efd1fb144336dc85ea04

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_11

The number $64$ has the property that it is divisible by its unit digit. How many whole numbers between 10 and 50 have this property? $\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 20$

Casework by the units digit $u$ will help organize the answer. $u=0$ gives no solutions, since no real numbers are divisible by $0$ $u=1$ has $4$ solutions, since all numbers are divisible by $1$ $u=2$ has $4$ solutions, since every number ending in $2$ is even (ie divisible by $2$ ). $u=3$ has $1$ solution: $33$ $\pm 10$ or $\pm 20$ will retain the units digit, but will stop the number from being divisible by $3$ $\pm 30$ is the smallest multiple of $10$ that will keep the number divisible by $3$ , but those numbers are $3$ and $63$ , which are out of the range of the problem. $u=4$ has $2$ solutions: $24$ and $44$ . Adding or subtracting $10$ will kill divisibility by $4$ , since $10$ is not divisible by $4$ $u=5$ has $4$ solutions: every number ending in $5$ is divisible by $5$ $u=6$ has $1$ solution: $36$ $\pm 10$ or $\pm 20$ will kill divisibility by $3$ , and thus kill divisibility by $6$ $u=7$ has no solutions. The first multiples of $7$ that end in $7$ are $7$ and $77$ , but both are outside of the range of this problem. $u=8$ has $1$ solution: $48$ $\pm 10, \pm 20, \pm 30$ will all kill divisibility by $8$ since $10, 20,$ and $30$ are not divisible by $8$ $u=9$ has no solutions. $9$ and $99$ are the smallest multiples of $9$ that end in $9$ Totalling the solutions, we have $0 + 4 + 4 + 1 + 2 + 4 + 1 + 0 + 1 + 0 = 17$ solutions, giving the answer $\boxed{17}$ , which is 17.

C

17

be1cb0eeaa4e8ebbe5db028eae5d532b

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_12

A block wall 100 feet long and 7 feet high will be constructed using blocks that are 1 foot high and either 2 feet long or 1 foot long (no blocks may be cut). The vertical joins in the blocks must be staggered as shown, and the wall must be even on the ends. What is the smallest number of blocks needed to build this wall? [asy] draw((0,0)--(6,0)--(6,1)--(5,1)--(5,2)--(0,2)--cycle); draw((0,1)--(5,1)); draw((1,1)--(1,2)); draw((3,1)--(3,2)); draw((2,0)--(2,1)); draw((4,0)--(4,1)); [/asy] $\text{(A)}\ 344\qquad\text{(B)}\ 347\qquad\text{(C)}\ 350\qquad\text{(D)}\ 353\qquad\text{(E)}\ 356$

Since the bricks are $1$ foot high, there will be $7$ rows. To minimize the number of blocks used, rows $1, 3, 5,$ and $7$ will look like the bottom row of the picture, which takes $\frac{100}{2} = 50$ bricks to construct. Rows $2, 4,$ and $6$ will look like the upper row pictured, which has $49$ 2-foot bricks in the middle, and $2$ 1-foot bricks on each end for a total of $51$ bricks. Four rows of $50$ bricks and three rows of $51$ bricks totals $4\cdot 50 + 3\cdot 51 = 200 + 153 = 353$ bricks, giving the answer $\boxed{353}.$

D

353

bc6071e2be1d8a1ef18d884220f682f5

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_13

In triangle $CAT$ , we have $\angle ACT =\angle ATC$ and $\angle CAT = 36^\circ$ . If $\overline{TR}$ bisects $\angle ATC$ , then $\angle CRT =$ [asy] pair A,C,T,R; C = (0,0); T = (2,0); A = (1,sqrt(5+sqrt(20))); R = (3/2 - sqrt(5)/2,1.175570); draw(C--A--T--cycle); draw(T--R); label("$A$",A,N); label("$T$",T,SE); label("$C$",C,SW); label("$R$",R,NW);[/asy] $\text{(A)}\ 36^\circ\qquad\text{(B)}\ 54^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 90^\circ\qquad\text{(E)}\ 108^\circ$

In $\triangle ACT$ , the three angles sum to $180^\circ$ , and $\angle C = \angle T$ $\angle CAT + \angle ATC + \angle ACT = 180$ $36 + \angle ATC + \angle ATC = 180$ $2 \angle ATC = 144$ $\angle ATC = 72$ Since $\angle ATC$ is bisected by $\overline{TR}$ $\angle RTC = \frac{72}{2} = 36$ Now focusing on the smaller $\triangle RTC$ , the sum of the angles in that triangle is $180^\circ$ , so: $\angle RTC + \angle TCR + \angle CRT = 180$ $36 + \angle ACT + \angle CRT = 180$ $36 + \angle ATC + \angle CRT = 180$ $36 + 72 + \angle CRT = 180$ $\angle CRT = 72^\circ$ , giving the answer $\boxed{72}$

C

72

40fdc756f0eaf527a5889f4d219b455d

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_15

Triangles $ABC$ $ADE$ , and $EFG$ are all equilateral. Points $D$ and $G$ are midpoints of $\overline{AC}$ and $\overline{AE}$ , respectively. If $AB = 4$ , what is the perimeter of figure $ABCDEFG$ [asy] pair A,B,C,D,EE,F,G; A = (4,0); B = (0,0); C = (2,2*sqrt(3)); D = (3,sqrt(3)); EE = (5,sqrt(3)); F = (5.5,sqrt(3)/2); G = (4.5,sqrt(3)/2); draw(A--B--C--cycle); draw(D--EE--A); draw(EE--F--G); label("$A$",A,S); label("$B$",B,SW); label("$C$",C,N); label("$D$",D,NE); label("$E$",EE,NE); label("$F$",F,SE); label("$G$",G,SE);[/asy] $\text{(A)}\ 12\qquad\text{(B)}\ 13\qquad\text{(C)}\ 15\qquad\text{(D)}\ 18\qquad\text{(E)}\ 21$

The large triangle $ABC$ has sides of length $4$ . The medium triangle has sides of length $2$ . The small triangle has sides of length $1$ . There are $3$ segment sizes, and all segments depicted are one of these lengths. Starting at $A$ and going clockwise, the perimeter is: $AB + BC + CD + DE + EF + FG + GA$ $4 + 4 + 2 + 2 + 1 + 1 + 1$ $15$ , thus the answer is $\boxed{15}$

C

15

40fdc756f0eaf527a5889f4d219b455d

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_15

Triangles $ABC$ $ADE$ , and $EFG$ are all equilateral. Points $D$ and $G$ are midpoints of $\overline{AC}$ and $\overline{AE}$ , respectively. If $AB = 4$ , what is the perimeter of figure $ABCDEFG$ [asy] pair A,B,C,D,EE,F,G; A = (4,0); B = (0,0); C = (2,2*sqrt(3)); D = (3,sqrt(3)); EE = (5,sqrt(3)); F = (5.5,sqrt(3)/2); G = (4.5,sqrt(3)/2); draw(A--B--C--cycle); draw(D--EE--A); draw(EE--F--G); label("$A$",A,S); label("$B$",B,SW); label("$C$",C,N); label("$D$",D,NE); label("$E$",EE,NE); label("$F$",F,SE); label("$G$",G,SE);[/asy] $\text{(A)}\ 12\qquad\text{(B)}\ 13\qquad\text{(C)}\ 15\qquad\text{(D)}\ 18\qquad\text{(E)}\ 21$

The perimeter of $ABCDEFG$ is the perimeter of the three triangles, minus segments $AD$ and $EG$ , which are on the interior of the figure. Because each of these segments is on two triangles, each segment must be subtracted two times. As in solution 1, the sides of the triangles are $4, 2,$ and $1$ , and the perimeters of the triangles are thus $12, 6,$ and $3$ The perimeter of the three triangles is $12 + 6 + 3 = 21$ . Subtracting the two segments $AD$ and $EG$ two times, the perimeter of $ABCDEFG$ is $21 - 2 - 1 - 2 - 1 = 15$ , and the answer is $\boxed{15}$

C

15

8d87abcc79e9ddb45e48d1366c6b1f6c

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_16

In order for Mateen to walk a kilometer (1000m) in his rectangular backyard, he must walk the length 25 times or walk its perimeter 10 times. What is the area of Mateen's backyard in square meters? $\text{(A)}\ 40 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 400 \qquad \text{(D)}\ 500 \qquad \text{(E)}\ 1000$

The length $L$ of the rectangle is $\frac{1000}{25}=40$ meters. The perimeter $P$ is $\frac{1000}{10}=100$ meters. Since $P_{rect} = 2L + 2W$ , we plug values in to get: $100 = 2\cdot 40 + 2W$ $100 = 80 + 2W$ $2W = 20$ $W = 10$ meters Since $A_{rect} = LW$ , the area is $40\cdot 10=400$ square meters or $\boxed{400}$

C

400

9483c1869b72efaa14b4e7e143bf5321

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_19

Three circular arcs of radius $5$ units bound the region shown. Arcs $AB$ and $AD$ are quarter-circles, and arc $BCD$ is a semicircle. What is the area, in square units, of the region? [asy] pair A,B,C,D; A = (0,0); B = (-5,5); C = (0,10); D = (5,5); draw(arc((-5,0),A,B,CCW)); draw(arc((0,5),B,D,CW)); draw(arc((5,0),D,A,CCW)); label("$A$",A,S); label("$B$",B,W); label("$C$",C,N); label("$D$",D,E);[/asy] $\text{(A)}\ 25\qquad\text{(B)}\ 10+5\pi\qquad\text{(C)}\ 50\qquad\text{(D)}\ 50+5\pi\qquad\text{(E)}\ 25\pi$

Draw two squares: one that has opposing corners at $A$ and $B$ , and one that has opposing corners at $A$ and $D$ . These squares share side $\overline{AO}$ , where $O$ is the center of the large semicircle. These two squares have a total area of $2 \cdot 5^2$ , but have two quarter circle "bites" of radius $5$ that must be removed. Thus, the bottom part of the figure has area $2\cdot 25 - 2 \cdot \frac{1}{4}\pi \cdot 5^2$ $50 - \frac{25\pi}{2}$ This is the area of the part of the figure underneath $\overline{BD}$ . The part of the figure over $\overline{BD}$ is just a semicircle with radius $5$ , which has area of $\frac{1}{2}\pi\cdot 5^2 = \frac{25\pi}{2}$ Adding the two areas gives a total area of $\boxed{50}$

50

79a4ab72d07510cff6b4f762be595a81

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_20

You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $ $1.02$ , with at least one coin of each type. How many dimes must you have? $\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$

Since you have one coin of each type, $1 + 5 + 10 + 25 = 41$ cents are already determined, leaving you with a total of $102 - 41 = 61$ cents remaining for $5$ coins. You must have $1$ more penny. If you had more than $1$ penny, you must have at least $6$ pennies to leave a multiple of $5$ for the nickels, dimes, and quarters. But you only have $5$ more coins to assign. Now you have $61 - 1 = 60$ cents remaining for $4$ coins, which may be nickels, quarters, or dimes. If you have only one more dime, that leaves $50$ cents in $3$ nickels or quarters, which is impossible. If you have two dimes, that leaves $40$ cents for $2$ nickels or quarters, which is again impossible. If you have three dimes, that leaves $30$ cents for $1$ nickel or quarter, which is still impossible. And all four remaining coins being dimes will not be enough. Therefore, you must have no more dimes to assign, and the $60$ cents in $4$ coins must be divided between the quarters and nickels. We quickly see that $2$ nickels and $2$ quarters work. Thus, the total count is $2$ quarters, $2$ nickels, $1$ penny, plus one more coin of each type that we originally subtracted. Double-checking, that gives a total $2 + 2 + 1 + 4 = 9$ coins, and a total of $2\cdot 25 + 2\cdot 5 + 1 + (1 + 5 + 10 + 25) = 102$ cents. There is only $1$ dime in that combo, so the answer is $\boxed{1}$

A

1

79a4ab72d07510cff6b4f762be595a81

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_20

You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $ $1.02$ , with at least one coin of each type. How many dimes must you have? $\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$

We see that there must be 102 cents, so therefore there's at least 2 pennies. That leaves 7 coins. We assume that there are 3 quarters, leaving 25 cents with 4 coins left. If all 4 are nickels, that would only be 20 cents, missing 5. Therefore, one nickel must be changed into 1 dime, so the answer is $\boxed{1}$

A

1

79a4ab72d07510cff6b4f762be595a81

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_20

You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $ $1.02$ , with at least one coin of each type. How many dimes must you have? $\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$

It is clear that there should only be $2$ pennies; any more would take up too many coins, and the limit is $9$ . Now we have $ $1$ left, and $7$ coins to use. Looking at the quarters, we can make methodical guesses. If there is $1$ quarter, then we would have to make $ $0.75$ with $6$ coins. We take a few educated guesses for the nickel and dime combos, and see that $1$ quarter will not work. Trying values for $2$ quarters, we see that this will not work either. When we reach $3$ quarters, the remaining is $ $0.25$ made from $4$ coins. We try with $2$ dimes, which does not work (it only takes $3$ coins) and we try with $1$ dime. After trying $2$ pennies, $3$ quarters, $1$ dime, and $3$ nickels, it is evident that this combo works. Therefore, the answer is $\boxed{1}$

A

1

2207a7ef9cb752c2d16de0327484533c

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_21

Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is $\text{(A)}\ \frac{1}{4}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{2}{3}\qquad\text{(E)}\ \frac{3}{4}$

Divide it into $2$ cases: 1) Keiko and Ephriam both get $0$ heads: This means that they both roll all tails, so there is only $1$ way for this to happen. 2) Keiko and Ephriam both get $1$ head: For Keiko, there is only $1$ way for this to happen because he is only flipping 1 penny, but for Ephriam, there are 2 ways since there are $2$ choices for when he can flip the head. So, in total there are $2 \cdot 1 = 2$ ways for this case. Thus, in total there are $3$ ways that work. Since there are $2$ choices for each coin flip (Heads or Tails), there are $2^3 = 8$ total ways of flipping 3 coins. Thus, since all possible coin flips of 3 coins are equally likely, the probability is $\boxed{38}$

B

38

8774cad6795f4ebd48043dc4258bd9c8

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_22

A cube has edge length $2$ . Suppose that we glue a cube of edge length $1$ on top of the big cube so that one of its faces rests entirely on the top face of the larger cube. The percent increase in the surface area (sides, top, and bottom) from the original cube to the new solid formed is closest to [asy] draw((0,0)--(2,0)--(3,1)--(3,3)--(2,2)--(0,2)--cycle); draw((2,0)--(2,2)); draw((0,2)--(1,3)); draw((1,7/3)--(1,10/3)--(2,10/3)--(2,7/3)--cycle); draw((2,7/3)--(5/2,17/6)--(5/2,23/6)--(3/2,23/6)--(1,10/3)); draw((2,10/3)--(5/2,23/6)); draw((3,3)--(5/2,3));[/asy] $\text{(A)}\ 10\qquad\text{(B)}\ 15\qquad\text{(C)}\ 17\qquad\text{(D)}\ 21\qquad\text{(E)}\ 25$

The original cube has $6$ faces, each with an area of $2\cdot 2 = 4$ square units. Thus the original figure had a total surface area of $24$ square units. The new figure has the original surface, with $6$ new faces that each have an area of $1$ square unit, for a total surface area of of $6$ additional square units added to it. But $1$ square unit of the top of the bigger cube, and $1$ square unit on the bottom of smaller cube, is not on the surface, and does not count towards the surface area. The total surface area is therefore $24 + 6 - 1 - 1 = 28$ square units. The percent increase in surface area is $\frac{SA_{new} - SA_{old}}{SA_{old}}\cdot 100\% = \frac{28-24}{24}\cdot 100\% \approx 16.67\%$ , giving the closest answer as $\boxed{17}$

C

17

6a0fcdbae0233f9585412de1189ec1c0

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_23

There is a list of seven numbers. The average of the first four numbers is $5$ , and the average of the last four numbers is $8$ . If the average of all seven numbers is $6\frac{4}{7}$ , then the number common to both sets of four numbers is $\text{(A)}\ 5\frac{3}{7}\qquad\text{(B)}\ 6\qquad\text{(C)}\ 6\frac{4}{7}\qquad\text{(D)}\ 7\qquad\text{(E)}\ 7\frac{3}{7}$

Remember that if a list of $n$ numbers has an average of $k$ , then the sum $S$ of all the numbers on the list is $S = nk$ So if the average of the first $4$ numbers is $5$ , then the first four numbers total $4 \cdot 5 = 20$ If the average of the last $4$ numbers is $8$ , then the last four numbers total $4 \cdot 8 = 32$ If the average of all $7$ numbers is $6\frac{4}{7}$ , then the total of all seven numbers is $7 \cdot 6\frac{4}{7} = 7\cdot 6 + 4 = 46$ If the first four numbers are $20$ , and the last four numbers are $32$ , then all "eight" numbers are $20 + 32 = 52$ . But that's counting one number twice. Since the sum of all seven numbers is $46$ , then the number that was counted twice is $52 - 46 = 6$ , and the answer is $\boxed{6}$

B

6

a970a4108aaf1205100810f1093e577e

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_24

If $\angle A = 20^\circ$ and $\angle AFG =\angle AGF$ , then $\angle B+\angle D =$ [asy] pair A,B,C,D,EE,F,G; A = (0,0); B = (9,4); C = (21,0); D = (13,-12); EE = (4,-16); F = (13/2,-6); G = (8,0); draw(A--C--EE--B--D--cycle); label("$A$",A,W); label("$B$",B,N); label("$C$",C,E); label("$D$",D,SE); label("$E$",EE,SW); label("$F$",F,WSW); label("$G$",G,NW);[/asy] $\text{(A)}\ 48^\circ\qquad\text{(B)}\ 60^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 80^\circ\qquad\text{(E)}\ 90^\circ$

As a strategy, think of how $\angle B + \angle D$ would be determined, particularly without determining either of the angles individually, since it may not be possible to determine $\angle B$ or $\angle D$ alone. If you see $\triangle BFD$ , then you can see that the problem is solved quickly after determining $\angle BFD$ But start with $\triangle AGF$ , since that's where most of our information is. Looking at $\triangle AGF$ , since $\angle F = \angle G$ , and $\angle A = 20$ , we can write: $\angle A + \angle G + \angle F = 180$ $20 + 2\angle F = 180$ $\angle AFG = 80$ By noting that $\angle AFG$ and $\angle GFD$ make a straight line, we know $\angle AFG + \angle GFD = 180$ $80 + \angle GFD = 180$ $\angle GFD = 100$ Ignoring all other parts of the figure and looking only at $\triangle BFD$ , you see that $\angle B + \angle D + \angle F = 180$ . But $\angle F$ is the same as $\angle GFD$ . Therefore: $\angle B + \angle D + \angle GFD = 180$ $\angle B + \angle D + 100 = 180$ $\angle B + \angle D = 80^\circ$ , and the answer is thus $\boxed{80}$

D

80

cfabdc2160ceb04a5d1d5c049abae0eb

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_25

The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is [asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW);[/asy] $\text{(A)}\ 21\qquad\text{(B)}\ 27\qquad\text{(C)}\ 30\qquad\text{(D)}\ 36\qquad\text{(E)}\ 40$

To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that $ABCD$ can have any dimension. Give the rectangle dimensions of $AB = CD = 12$ and $BC = AD= 6$ , which is the easiest way to avoid fractions. Labelling the right midpoint as $M$ , and the bottom midpoint as $N$ , we know that $DN = NC = 6$ , and $BM = MC = 3$ $[\triangle ADN] = \frac{1}{2}\cdot 6\cdot 6 = 18$ $[\triangle MNC] = \frac{1}{2}\cdot 3\cdot 6 = 9$ $[\triangle ABM] = \frac{1}{2}\cdot 12\cdot 3 = 18$ $[\triangle AMN] = [\square ABCD] - [\triangle ADN] - [\triangle MNC] - [\triangle ABM]$ $[\triangle AMN] = 72 - 18 - 9 - 18$ $[\triangle AMN] = 27$ , and the answer is $\boxed{27}$

B

27

cfabdc2160ceb04a5d1d5c049abae0eb

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_25

The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is [asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW);[/asy] $\text{(A)}\ 21\qquad\text{(B)}\ 27\qquad\text{(C)}\ 30\qquad\text{(D)}\ 36\qquad\text{(E)}\ 40$

The above answer is fast, but satisfying, and assumes that the area of $\triangle AMN$ is independent of the dimensions of the rectangle. All in all, it's a very good answer though. However this is an alternative if you don't get the above answer. Label $AB = CD = l$ and $BC = DA = h$ Labelling $m$ and $n$ as the right and lower midpoints respectively, and redoing all the work above, we get: $[\triangle ABN] = \frac{1}{2}\cdot h\cdot \frac{l}{2} = \frac{lh}{4}$ $[\triangle MNC] = \frac{1}{2}\cdot \frac{l}{2}\cdot \frac{w}{2} = \frac{lh}{8}$ $[\triangle ABM] = \frac{1}{2}\cdot l\cdot \frac{h}{2} = \frac{lh}{4}$ $[\triangle AMN] = [\square ABCD] - [\triangle ADN] - [\triangle MNC] - [\triangle ABM]$ $[\triangle AMN] = lh - \frac{lh}{4} - \frac{lh}{8} - \frac{lh}{4}$ $[\triangle AMN] = \frac{3}{8}lh = \frac{3}{8}\cdot 72 = 27$ , and the answer is $\boxed{27}$

B

27

cfabdc2160ceb04a5d1d5c049abae0eb

https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_25

The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is [asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW);[/asy] $\text{(A)}\ 21\qquad\text{(B)}\ 27\qquad\text{(C)}\ 30\qquad\text{(D)}\ 36\qquad\text{(E)}\ 40$

Let's assume, for simplicity, that the sides of the rectangle are $9$ and $8.$ The area of the 3 triangles would then be $8\cdot\frac{9}{2}\cdot\frac{1}{2} = 18,$ $4\cdot\frac{9}{2}\cdot\frac{1}{2} = 9,$ $4\cdot 9\cdot\frac{1}{2} = 18.$ Adding these up, we get $45$ , and subtracting that from $72$ , we get $27$ , so the answer is $\boxed{27}$

B

27

80bf832f3e22e1392ce1190812caa675

https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_2

What is the degree measure of the smaller angle formed by the hands of a clock at 10 o'clock? [asy] draw(circle((0,0),2)); dot((0,0)); for(int i = 0; i < 12; ++i) { dot(2*dir(30*i)); } label("$3$",2*dir(0),W); label("$2$",2*dir(30),WSW); label("$1$",2*dir(60),SSW); label("$12$",2*dir(90),S); label("$11$",2*dir(120),SSE); label("$10$",2*dir(150),ESE); label("$9$",2*dir(180),E); label("$8$",2*dir(210),ENE); label("$7$",2*dir(240),NNE); label("$6$",2*dir(270),N); label("$5$",2*dir(300),NNW); label("$4$",2*dir(330),WNW); [/asy] $\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 60 \qquad \textbf{(D)}\ 75 \qquad \textbf{(E)}\ 90$

At $10:00$ , the hour hand will be on the $10$ while the minute hand on the $12$ This makes them $\frac{1}{6}$ th of a circle apart, and $\frac{1}{6}\cdot360^{\circ}=\boxed{60}$

C

60

80bf832f3e22e1392ce1190812caa675

https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_2

What is the degree measure of the smaller angle formed by the hands of a clock at 10 o'clock? [asy] draw(circle((0,0),2)); dot((0,0)); for(int i = 0; i < 12; ++i) { dot(2*dir(30*i)); } label("$3$",2*dir(0),W); label("$2$",2*dir(30),WSW); label("$1$",2*dir(60),SSW); label("$12$",2*dir(90),S); label("$11$",2*dir(120),SSE); label("$10$",2*dir(150),ESE); label("$9$",2*dir(180),E); label("$8$",2*dir(210),ENE); label("$7$",2*dir(240),NNE); label("$6$",2*dir(270),N); label("$5$",2*dir(300),NNW); label("$4$",2*dir(330),WNW); [/asy] $\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 60 \qquad \textbf{(D)}\ 75 \qquad \textbf{(E)}\ 90$

We know that the full clock is a circle, and therefore has 360 degrees. Considering that there are $12$ numbers, the distance between one number will be $360\div 12=30$ . If the time is $10:00$ , then the hour hand will be on $10$ , and the minute hand will be on, $12$ , making them $2$ numbers apart, so they will be $60$ degrees apart, or $\boxed{60}$

C

60

3e181217a6b1cf3b680e0860595e489d

https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_4

The diagram shows the miles traveled by bikers Alberto and Bjorn. After four hours, about how many more miles has Alberto biked than Bjorn? [asy] for (int a = 0; a < 6; ++a) { for (int b = 0; b < 6; ++b) { dot((4*a,3*b)); } } draw((0,0)--(20,0)--(20,15)--(0,15)--cycle); draw((0,0)--(16,12)); draw((0,0)--(16,9)); label(rotate(30)*"Bjorn",(12,6.75),SE); label(rotate(37)*"Alberto",(11,8.25),NW); label("$0$",(0,0),S); label("$1$",(4,0),S); label("$2$",(8,0),S); label("$3$",(12,0),S); label("$4$",(16,0),S); label("$5$",(20,0),S); label("$0$",(0,0),W); label("$15$",(0,3),W); label("$30$",(0,6),W); label("$45$",(0,9),W); label("$60$",(0,12),W); label("$75$",(0,15),W); label("H",(6,-2),S); label("O",(8,-2),S); label("U",(10,-2),S); label("R",(12,-2),S); label("S",(14,-2),S); label("M",(-4,11),N); label("I",(-4,9),N); label("L",(-4,7),N); label("E",(-4,5),N); label("S",(-4,3),N); [/asy] $\text{(A)}\ 15 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 25 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$

After 4 hours, we see that Bjorn biked 45 miles, and Alberto biked 60. Thus the answer is $60-45=15$ $\boxed{15}$

A

15

3e181217a6b1cf3b680e0860595e489d

https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_4

The diagram shows the miles traveled by bikers Alberto and Bjorn. After four hours, about how many more miles has Alberto biked than Bjorn? [asy] for (int a = 0; a < 6; ++a) { for (int b = 0; b < 6; ++b) { dot((4*a,3*b)); } } draw((0,0)--(20,0)--(20,15)--(0,15)--cycle); draw((0,0)--(16,12)); draw((0,0)--(16,9)); label(rotate(30)*"Bjorn",(12,6.75),SE); label(rotate(37)*"Alberto",(11,8.25),NW); label("$0$",(0,0),S); label("$1$",(4,0),S); label("$2$",(8,0),S); label("$3$",(12,0),S); label("$4$",(16,0),S); label("$5$",(20,0),S); label("$0$",(0,0),W); label("$15$",(0,3),W); label("$30$",(0,6),W); label("$45$",(0,9),W); label("$60$",(0,12),W); label("$75$",(0,15),W); label("H",(6,-2),S); label("O",(8,-2),S); label("U",(10,-2),S); label("R",(12,-2),S); label("S",(14,-2),S); label("M",(-4,11),N); label("I",(-4,9),N); label("L",(-4,7),N); label("E",(-4,5),N); label("S",(-4,3),N); [/asy] $\text{(A)}\ 15 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 25 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$

We see that each dot is $15$ units away from the nearest one above it. So the answer is $\boxed{15}$

A

15

93ce2746c257aa6756cd4b9f90f08098

https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_5

A rectangular garden 60 feet long and 20 feet wide is enclosed by a fence. To make the garden larger, while using the same fence, its shape is changed to a square. By how many square feet does this enlarge the garden? $\text{(A)}\ 100 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 300 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 500$

We need the same perimeter as a $60$ by $20$ rectangle, but the greatest area we can get. right now the perimeter is $160$ . To get the greatest area while keeping a perimeter of $160$ , the sides should all be $40$ . that means an area of $1600$ . Right now, the area is $20 \times 60$ which is $1200$ $1600-1200=400$ which is $\boxed{400}$

D

400

1a45145db57d4dc89dec6a41c8d24788

https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_7

The third exit on a highway is located at milepost 40 and the tenth exit is at milepost 160. There is a service center on the highway located three-fourths of the way from the third exit to the tenth exit. At what milepost would you expect to find this service center? $\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 110 \qquad \text{(D)}\ 120 \qquad \text{(E)}\ 130$

There are $160-40=120$ miles between the third and tenth exits, so the service center is at milepost $40+(3/4)(120) = 40+90=\boxed{130}$

E

130

1012fac17ceea4a2d3ce80fa557cee0c

https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_9

Three flower beds overlap as shown. Bed A has 500 plants, bed B has 450 plants, and bed C has 350 plants. Beds A and B share 50 plants, while beds A and C share 100. The total number of plants is [asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw(circle((.3,-.1),.7)); draw(circle((2.8,-.2),.8)); label("A",(1.3,.5),N); label("B",(3.1,-.2),S); label("C",(.6,-.2),S); [/asy] $\text{(A)}\ 850 \qquad \text{(B)}\ 1000 \qquad \text{(C)}\ 1150 \qquad \text{(D)}\ 1300 \qquad \text{(E)}\ 1450$

Plants shared by two beds have been counted twice, so the total is $500 + 450 + 350 - 50 - 100 = \boxed{1150}$

C

1150

1012fac17ceea4a2d3ce80fa557cee0c

https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_9

Three flower beds overlap as shown. Bed A has 500 plants, bed B has 450 plants, and bed C has 350 plants. Beds A and B share 50 plants, while beds A and C share 100. The total number of plants is [asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw(circle((.3,-.1),.7)); draw(circle((2.8,-.2),.8)); label("A",(1.3,.5),N); label("B",(3.1,-.2),S); label("C",(.6,-.2),S); [/asy] $\text{(A)}\ 850 \qquad \text{(B)}\ 1000 \qquad \text{(C)}\ 1150 \qquad \text{(D)}\ 1300 \qquad \text{(E)}\ 1450$

Bed A has $350$ plants it doesn't share with B or C. Bed B has $400$ plants it doesn't share with A or C. And C has $250$ it doesn't share with A or B. The total is $350 + 400 + 250 + 50 + 100 = \boxed{1150}$ plants.

C

1150